I'll modify my previous answer, then will apply it to derive the Post algebra as presented in @Emil's comment above.

Let $\ T\ $ be an arbitrary finite $n$-element set ($n\ge 2$). We may label it so that it becomes $\ T=\{0\ldots n\!-\!1\}$, and let $\ \Lambda:=n-1$. Now let $\ \wedge\ $ and $\ \vee\ $ be short for $ \min \ $ and $\ \max\ $ respectively, and $\ \bigwedge\ $ and $\ \bigvee\ $ be their finite iterations. Also let $\ \lambda_a : T\rightarrow T\ $ be defined by

$$ \lambda_a(x)\ :=\ \Lambda\quad\Leftrightarrow\quad a=x$$
$$ \lambda_a(x)\ :=\ 0\quad\Leftrightarrow\quad a\ne x$$

for every $\ x\in T$. Consider arbitrary $D$-argument operation $\ f:T^D\rightarrow T\ $, where $\ D\ $ is a finite set. Then

$$ f\ =\ \bigvee_{\tau\in T^D}\ (\,f(\tau)\ \wedge\ \bigwedge_{d\in D}(\lambda_{\tau(d)}\circ\pi_d)\,)$$

where $\ \pi_d:T^D\rightarrow T\ $ is the cartesian projection for every $\ d\in D$.

Thus we have a (modified) complete set of operations--it consists of $\ n\ $ constants, $\ n\ $ comparisons, and $\ \wedge\ $, and $\ \vee$.

Now let's show that the Post set $\ \{\vee\ \ \neg\ \}\ $ is complete.

## Completeness of $\ \{\vee\ \ \neg\ \}$

(The equalities below are theorems, not definitions).

First of all we get constant $\ \Lambda$; it is the maximum of the consecutive compositions of the negation:

$$\Lambda\ =\ \bigvee_{k=0}^{n-1} \bigcirc^k\neg$$

Of course we identify the constants and the constant operations. Now we get every other constant $\ c\in T$:

$$ c\ = \left(\bigcirc^{c+1} \neg\right)\circ\Lambda$$

for every $\ c\in T$.

Now it's time to obtain the comparisons. Let's start with:

$$ \lambda_0\ \ =\ \ \neg\ \circ\ \bigvee_{k=0}^{n-2} \bigcirc^k\neg$$

Next:

$$ \forall_{a\in T}\quad \lambda_a\ =\ \lambda_0\,\circ\,\bigcirc^{n-a}\,\neg$$

To complete the Post's completeness only $\ \wedge\ $ is left. But first let's define an order negation $\ \sim\,:T\rightarrow T\ $ to the Post's algebraic negation $\ \neg\,$:

$$ \sim\ \ =\ \ \bigvee_{a=0}^{n-1}\ \left(\left(\bigcirc^{n-a}\ \neg\right)\,\circ\,\lambda_a\right) $$

Now the last touch is provided by the De Morgan law:

$$x\,\wedge\,y\ =\ \sim\left(\,\sim\!(x)\ \vee\,\sim\!(y)\,\right)$$

**REMARK** One could use exponent $\,\ -\!a\ \,$ instead of $\,\ n-a$.

allfunctions on a set, it is irrelevant what field structure you chose to endow the set with. Am I missing something? $\endgroup$fields(besides constraining the size of $X$ to being a prime power, which turns out not to make any difference), it is just about functions on finitesets. Since any characterization of functionally complete sets of functions must be invariant under permutations of $X$, it will not respect any additional algebraic structure (like field operations) which you may put on the set. $\endgroup$1more comment